Respuesta :
Answer:
The probability that exactly two are defective given that the number of defective bulbs is two or fewer is 0.4040.
Step-by-step explanation:
Let X be the number of defective bulbs.
[tex]X\sim B(n,p)[/tex]
Where n is sample size and p is probability of success.
According to the given information,
[tex]n=100[/tex]
[tex]p=0.02[/tex]
[tex]q=1-p=1-0.02=0.98[/tex]
According to binomial distribution the probability of exactly r success from n is
[tex]P(X=r)=^nC_rp^rq^{n-r}[/tex]
The probability that exactly two are defective is
[tex]P(X=2)=^{100}C_2(0.02)^2(0.98)^{98}\approx 0.2724[/tex]
The probability that the number of defective bulbs is two or fewer is
[tex]P(X\leq 2)=P(X=0)+P(X=1)+P(X=2)[/tex]
[tex]P(X\leq 2)=^{100}C_0(0.02)^0(0.98)^{100}+^{100}C_1(0.02)^1(0.98)^{99}+^{100}C_2(0.02)^2(0.98)^{98}[/tex]
[tex]P(X\leq 2)=0.1326+0.2707+0.2734[/tex]
[tex]P(X\leq 2)=0.6767[/tex]
We have to find the probability that exactly two are defective given that the number of defective bulbs is two or fewer.
[tex]P(x=2|x\leq 2)[/tex]
According to the conditional probability
[tex]P(\frac{A}{B})=\frac{P(A\cap B)}{P(B)}[/tex]
[tex]P(x=2|x\leq 2)=\frac{P(x=2\cap x\leq 2)}{P(x\leq 2)}[/tex]
[tex]P(x=2|x\leq 2)=\frac{P(x=2)}{P(x\leq 2)}[/tex]
[tex]P(x=2|x\leq 2)=\frac{0.2734}{0.6767}=0.4040[/tex]
Therefore the probability that exactly two are defective given that the number of defective bulbs is two or fewer is 0.4040.
